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If $x$ is real, the set of real values of $a$ for which the function $$y=x^2-ax+1-2a^2$$ is always greater than zero is

  1. $- \frac{2}{3} < a \leq \frac{2}{3}$
  2. $- \frac{2}{3} \leq  a < \frac{2}{3}$
  3. $- \frac{2}{3} < a < \frac{2}{3}$
  4. None of these
in Calculus by Veteran (431k points)
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I have solved this by finding Discriminant.

and getting $x = -2a, \;\text {and} \;2a$

Is this the right method?

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